NORD - InformationCalculation Methods and Examples

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Physical Formulae

Linear motion Rotating motion

Distance s = v ∗ t Angular ϕ = ω ∗ t

Velocity (speed) v = st

Angular velocity ω = ϕτ = 2 ∗ π ∗ n

Accelleration a = vt

Angular accelleration α = Wt

Force F = m ∗ a Torque M = J ∗ r = F ∗ r

Power P = F ∗ v Power P = M ∗ ωWork W = F ∗ s = P ∗ t Work W = M ∗ ϕ = P ∗ t

Kinetic energy Wkin = 12 ∗ m ∗ v2 Rotating energy Wrot =

12 ∗ J ∗ ω2

Formulae of drive engineering

Rolling resistance, -force FR = m ∗ g ∗ 2D

∗ (µL ∗ d2 + f) + c

: µL, f, c, s. tables_____ _1_,_ _2_,_ _3_._

or FR = We ∗ m We for wheel / rail steel

s. diagramme 1, 2.

Sliding resitance, -force FG = m ∗ g ∗ µ µ s. table 4

Static friction-force FH = m ∗ g ∗ µO µO s. table 4

Windload F = A ∗ PW

Moment of inertia with refernce Jred = 91,2 ∗ m ∗

vnM

2 Translation

to the motor shaft or Jred = J ∗

nnM

2 Rotation

Speed n = v ∗ 60π ∗ D

Torque M = P ∗ 9550

n

Friction power PR = F ∗ v

1000 ∗ ηTranslation

or PR = M ∗ n9550

Rotation

Acceleration power PB = m ∗ a ∗ v1000 ∗ η Translation

or PB = J ∗ n2

91,2 ∗ 1000 ∗ tB ∗ η Rotation

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Hubleistung PHub = m ∗ g ∗ v

1000 ∗ η

Beschleunigung aB = 9,55 ∗ v ∗ (MH ± ML)

(Jred + JM + JBre + JZ) ∗ n

Beschleunigungszeit tB = vaB

Beschleunigungsweg sB = v2

2 ∗ aB

Verzögerung av = 9,55 ∗ v ∗ (MB ± ML)

(Jred + JM + JBre + JZ) ∗ n

Verzögerungszeit tV = vaV

Verzögerungsweg sV = v2

2 ∗ aV

zulässige Schalthäufigkeit zzul = 1 − ML ⁄ MH

1 + (Jred + JBre + JZ) ⁄ JM ∗ zo

Positioniergenauigkeit Positioniergenauigkeit = ± 0,25 * sv

Bremsarbeit WB = (Jred + JM + JBre + JZ) ∗ n2

182,5 ∗

MB

MB ± ML

Lebensdauer der Bremsbeläge LN = Wzul

WB ∗ z

Übersetzung i = n1

n2 =

M2

M1 =

d2

d1 =

z2

z1

Wirkungsgrad η = Pab

Pzuη s. Tabelle 5

rücktreibender Wirkungsgrad η G’ = 2 − 1

η G

Querkraft FQ = 2 ∗ M2

D ∗ fZ ∗ ≤ FQzul fZ s. Tabelle 6

Betriebsfaktor fB = M2max

M2

Massenbeschleunigungsfaktor maf = Jred

JM + JZ + JBre

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a

aB

av

A

c

d

dO

d1

d2

D

f

fBfZ

F

FG

FH

FQ

FQvorh

FQzul

FR

FW

g

i

iV

J

JBre

JM

Jred

JZ

LN

m

maf

mG

mL

mO

Formulae symbols and unities

Acceleration

Acceleration (start up)

Deceleration (braking)

Area (wind)

Additional factor for secondary friction

Diameter (bearing spigot diameter)

Pinion or sprocket diameter

Pinion diameter

Chain sprocket diameter

Diameter of the travelling wheel or cable drum or of thesprocket

Lever arm of rolling friction

Service factor

Additional factor for overhung load

Force, rolling resistance

Sliding friction

Static friction

Overhung load

Existing overhung load

Permissible overhung load

Rolling resistance

Wind load

Gravity (constant: 9,81)

Reduction

Additional reduction (gear, chain, belt ...)

Moment of inertia

Moment of inertia of the brake

Moment of inertia of the motor

Moment of inertia with reference to the motorshaft

Moment of inertia of the z-fan

Brake service life until readjustment

Weight (mass)

Inertia mass acceleration factor

Mass of counter weight

Mass with full load

Mass without load

m/s2

m/s2

m/s2

m2

-

m

m

m

m

m

m

-

-

N

N

N

N

N

N

N

N

m/s2

-

-

kgm2

kgm2

kgm2

kgm2

kgm2

h

kg

-

kg

kg

kg

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M

MB

MH

ML

MN

M1

M2

M2max

n

nM

nN

n1

n2

PW

P

Pab

Pzu

PB

PHub

PN

PR

r

s

sB

sV

t

tB

tV

v

W

We

Wkin

Wrot

Wzul

WB

x

Torque

Braking torque

Run up torque

Torque with full load (with reference to the motor shaft)

Rated torque

Input torque

Output torque

Maximum permissible output torqe

Speed

Motor speed

Rated speed

Input speed

Output speed

Wind pressure

Power

Required power

Supplied power

Acceleration power

Lifting power

Rated power

Fricition power

Radius

Distance

Start up distance

Braking distance

Time

Start up time

Braking time

Velocity (speed)

Work

Standard rolling friction

Kinetic energy

Rotating energy

Braking work until readjustment

Braking work

Number of drives

Nm

Nm

Nm

Nm

Nm

Nm

Nm

Nm

1/min1/min1/min1/min1/min

N/m2

kW

kW

kW

kW

kW

kW

kW

m

m

m

m

s

s

s

m/s

J

N/t

J

J

J

J

-

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z

zzul

zO

z1

z2

αηηG

ηG’

µµL

µO

ϕω

Starting frequency

Permissible starting frequency

Starting frequency with no load

Number of gear teeth pinion

Number of gear teeth gear wheel

Angular acceleration

Efficiency

Efficiency of gear unit

Reverse operating efficiency

Coefficient of friction

Coefficient of friction for bearings

Coefficient of friction (static)

Angular

Angular velocity

s/h

s/h

s/h

-

-

1/s2

-

-

-

-

-

-

°1/s

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table 1: coefficient of friction for bearings µL

table 2: lever arm of rolling friction f

table 3: rim friction on the wheels c

table 4: static friction and sliding friction µ

table 5: efficiency η

table 6: additional factor inderming overhung loads fz

Friction bearing Sliding bearings

µL 0,005 0,1

7

f

steel / steel 0,0005 m

wood / steel 0,0012 m

polymer / steel 0,002 m

hardrubber / steel 0,0077 m

hardrubber / concrete 0,01 - 0,02 m

rubber / concrete 0,015 - 0,035 m

Friction ofanti friction bearings

Friction ofsleeve bearings

Friction of guide rollers

c 0,003 0,005 0,002

Static friction µO Sliding friction µdry greased dry greased

steel / steel 0,11 - 0,40 0,10 0,10 - 0,30 0,01 - 0,10

steel / last iron 0,18 - 0,25 0,10 0,16 - 0,25 0,05 - 0,10

steel / wood 0,50 - 0,70 0,10 0,20 - 0,50 0,02 - 0,10

steel / polymer 0,20 - 0,50 0,10 - 0,35

steel / rubber 0,40 - 0,50

wood / wood 0,40 - 0,80 0,16 0,20 - 0,50 0,04 - 0,16

fZ

helical gears 1,1 z = 17 teeth

chain sprockets 1,4 z = 13 teeth

chain sprockets 1,2 z = 20 teeth

pulleys 1,7 by tensioning influence

pulleys 2,5 by tensioning influence

ηchain 0,90 - 0,96 per complete wrap of the rope around the drum

wire ropes 0,90 - 0,95 per complete wrap

flat polymer belts 0,93 - 0,98 per complete wrap of the rope depending on the material

V-belts 0,85 - 0,95 per complete wrap

rubber belts 0,80 - 0,85 per complete wrap

polymer belts 0,80 - 0,85 per complete wrap

helical inline gear 0,95 - 0,98 oil lubricated depending on the number of the stages

worm gear 0,30 - 0,93 oil lubricated depending on the number of starts of the worm

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Example I.1: Drive arrangement for crane

Mass without load of the crane mO 13800 kg

Mass without load of the mk 1800 kg

Load mL 15000 kg

Velocity v 0,17 / 0,66 m/s = 10/40 m/min

Diameter of the travelling wheel D 0,4 m

Number of drives x 2

Additional reduction iv 4,24

Mounting position B 3

Switching frequencies z 60 s/h

Efficiency η 0,85

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Motor arrangements

Standard rolling friction We

We = WO + 30 N/t W0 = 36 N/t s. diagram

We = 36 N/t + 30 N/t = 66 N/t 30 N/t additional for rim friction

Power P (at maximum velocity)

P = We ∗ m ∗ v1000 ∗ η

PO = 66 N ⁄ t ∗ (13,8 t + 1,8 t) ∗ 0,66 m⁄s

1000 ∗ 0,85 = 0,80 kW (without load)

PL = 66 N ⁄ t ∗ (13,8 t + 1,8 t + 15,0 t) ∗ 0,66 m⁄s

1000 ∗ 0,85 = 1,57 kW (with load)

Pmax = PL

2 ∗

mO + 2 ∗ (mK + mL )mO + mK + mL

= 1,57 kW

2 ∗

13,8 t + 2 ∗ (1,8 t + 15 t)13,8 t + 1,8 t + 15 t

Pmax = 1,22 kW (one-sided trolley)

Motor data

Type 100 L/80-20 WU Bre16 Z (2 pieces)

Rated output power PN 0,55 / 2,2 kW

Rated speed nN 670 / 2740 1/min

Rated torque MN 7,8 / 7,7 Nm

Permissible no-load starting frequency zo 4000 / 1400 s/h

Motor moment of inertia JM 0,0060 kgm2

Additonal moment of inertia Jz 0,0113 kgm2

Brake moment of inertia JBre 0,0001 kgm2

Braking torque MB (brake 16 adjusted to 8 Nm ) 8 Nm

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Gear arrangment

Wheel speed n L

nL = v ∗ 60π ∗ D

nL = 0,66 m⁄s ∗ 60

π ∗ 0,4 m = 32 1⁄min

Gear output speed n 2

n2 = nL ∗ iv

n2 = 32 1⁄min ∗ 4,24 = 136 1⁄min

Acceleration factor of mass m af

maf = Jred

JM + Jz + JBre

maf = 0,0810 kgm2

0,0060 kgm2 + 0,0113 kgm2 + 0,0001 kgm2 = 4,7

Starting freqeuency per hour: 180 (60 times acceleration, switching, deceleration)

⇒ Type of load C, fB = 1,6

Output torque M a

Ma = PN ∗ 9550

n2 ∗ fB

Ma = 2,2 kW ∗ 9550

136 1⁄min ∗ 1,6 = 247 Nm

For service factor fB = 1,6 the output torque of the gear is 247 Nm.

Reduction i

i = nN

n2

i = 2740 1⁄min

136 1⁄min = 20

Complete type: SK 22-100 L/80-20 WU Bre 16 ZPN = 0,55 / 2,2 kW

i = 20,03

n2 = 33 / 137 1/min

Mounting position B 3

Shaft ø 30 x 60 mm

Brake 16 Nm adjusted to 8 Nm

Special provision: special rotor

high inertia fan

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Example I.2: Drive arrangement for a trolley

Mass without load mk 1800 kg

Load mL 15000 kg

Velocity v 0,08 / 0,33 m/s = 5/20 m/min

Wheel diameter D 0,3 m

Number of drives x 1

Additional reduction iv 4

Mounting position B 5

Switching frequency z 60 s/h

Efficiency n 0,85

Pairing of material steel / steel

Guiding rim friction

Type of bearings (4 wheels) antifriction bearings

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Drive resistance:

FR = m ∗ g

2D

∗ ( µ L ∗ d2

+ f ) + c

Fro = 1800 kg ∗ 9,81 m

s2

20,3 m

∗ (0,005 ∗ 0,06

2 + 0,0005 m ) + 0,003m

Fro = 129,5 N ( without load)

FRL = 16800 kg ∗ 9,81 ms2

∗

20,3 m

(0,005 ∗ 0,06

2 m + 0,0005 m ) + 0,003 m

FRL = 1208,6 N (with load)

Power P (calculation for 2-poles gearmotors)

P = F ∗ V

1000 ∗ η

Po = 129,5 N ∗ 0,33 m

1000 ∗ 0,85 s = 0,05 kW

PL = 1208,6 N ∗ 0,33

1000 ∗ 0,85 = 0,47 kW

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Motor arrangement

Motor data

Type 100 L/8-2 WU Bre10 Z

Rated output power PN 0,4 / 1,6 kW

Rated speed nN 670 / 2740 1/min

Rated torque MN 5,7 / 5,6 Nm

Hochlaufmoment MH 9,2 / 8,6 Nm

No-load switching frequency zo 4200 / 1500 s/h

Motor moment of inertia JM 0,0045 kgm2

Moment of high inertia fan Jz 0,0113 kgm2

Brake moment of inertia JBre 0,0001 kgm2

Braking torque MB (brake 10 adjusted on 6 Nm) 6 Nm

Load torque M

M = P ∗ 9550x ∗ nN

MO = 0,05 KW ∗ 9550

2740 1⁄min = 0,2 Nm (without load)

ML = 0,47 kW ∗ 9550

2740 1⁄min = 1,6 Nm (with load)

Reduced moment of inertia J red

Jred = 1x

∗ 91,2 ∗ m ∗

vnN

2

JredO = 91,2 ∗ 1800 kg ∗

0,33 m⁄s2740 1⁄min

2

= 0,0024 kgm2

JredL = 91,2 ∗ 16800 kg ∗

0,33 m⁄s2740 1⁄min

2

= 0,0222 kgm2

Acceleration aB

aB = 9,55 ∗ v ∗ (MH − ML)

(Jred ⁄ η + JM + JBre + JZ) ∗ nN

aB = 9,55 ∗ 0,33 m⁄s ∗ (8,6 Nm − 0,2 Nm)

(0,0024 kgm2 ⁄ 0,85 + 0,0045 kgm2 + 0,0001 kgm2) ∗ 2740 1⁄min = 0,52 m⁄s2 (without load)

aB = 9,55 ∗ 0,33 m⁄s ∗ (8,6 Nm − 1,6 Nm)

(0,0222 kgm2 ⁄ 0,85 + 0,0045 kgm2 + 0,0001 kgm2) ∗ 2740 1⁄min = 0,19 m⁄s2 (with load)

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Decceleration a V

av = 9,55 ∗ v ∗ (MB + ML ∗ η2)

(Jred ∗ η + JM + JBre + JZ) ∗ nN

aVO = 9,55 ∗ 0,08 m⁄s ∗ (6 Nm + 0,2 Nm ∗ 0,852)

(0,0024 kgm2 ∗ 0,85 + 0,0045 kgm2 + 0,0001 kgm2 + 0,0113 kgm2) ∗ 670 1⁄min = 0,39 m⁄s2 (without load)

aVL = 9,55 ∗ 0,08 m⁄s ∗ (6 Nm + 1,6 Nm ∗ 0,852)

(0,0222 kgm2 ∗ 0,85 + 0,0045 kgm2 + 0,0001 kgm2 + 0,0113 kgm2) ∗ 670 1⁄min = 0,24 m⁄s2 (with load)

Permissible switching fr equency z zul

zzul = 1 − ML ⁄ MH

1 + (Jred + JZ + JBre) ⁄ JM ∗ z0

zzul = 1 − 1,6 Nm ⁄ 8,6 Nm

1 +(0,0222 kgm2 + 0,0113 kgm2 + 0,0001 kgm2) ⁄ 0,0045 kgm2 ∗ 1500 s⁄h = 142 s⁄h

The perm. switching frequency is calculated for the acceptance: starting 2-pole with load (every time) is not correct

because of delay for switching and running on the 8-pole.

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Gear arrangements

Wheel speedl nL

nL = v ∗ 60π ∗ D

nL = 0,33 m⁄s ∗ 60

π ∗ 0,3 m = 21 1⁄min

Gear unit output speed n 2

n2 = nL ∗ iv

n2 = 21 1⁄min ∗ 4 = 84 1⁄min

Mass acceleration factor maf

maf = Jred

JM + Jz + JBre

maf = 0,022 kgm2

0,0045 kgm2 + 0,0113 kgm2 + 0,0001 kgm2 = 1,4

Circuit m per hour: 180 (each 60 accelerations, switching, decelerations)

⇒ type of load B

⇒ fB ≥ 1,3

Output torque M a

Ma = PN ∗ 9550

n2 ∗ fB

Ma = 1,6 kW ∗ 9550

84 1⁄min ∗ 1,3 = 236 Nm

For service factor fB = 1,3 the output torque of the gear is 236 Nm.

Reduction i

i = nN

n2

i = 2740 1⁄min

84 1⁄min = 33

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Complete type: SK 22 F - 100 L/8-2 WU Bre 10 Z

PN = 0,4 / 0,16 kW

i = 34,69

n2 = 19/79 1/min

Mounting position B 5

Shaft ø 30 x 60 mm

Flange ø 160 mm oder 200 mm

Brake 10 Nm adjusted on 6 Nm

Special provision: special rotor (WU-silumin rotor)

high inertia fan

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NORD - Information

Example II.1: Drive unit for vert ical motion

Mass without load mO 50 kg

Load mL 200 kg

midle drum diameter Dm 0,208 m

Max. lifting speed v 0,24 m/s = 14,4 m/min

Operation cycle 8 h/Tag, 40 % ED

Starting frequency z 360 Hubbewegungen/h

Efficiency η 0,8

Positioning accuracy ± 1 mm

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Motor arrangement

Power P

P = m ∗ g ∗ v1000 ∗ η

PL = (50 kg + 200 kg) ∗ 9,81 m⁄s2 ∗ 0,24 m⁄s

1000 ∗ 0,8 = 0,74 kW

To get the required accuracy of ± 1 mm we have to choose a polechanging motor.

Motor data

Typ 80 L/4-2 Bre8

Rated output power PN 0,60 / 0,75 kW

Rated speed nN 1400 / 2830 1/min

Synchronous speed nsyn 1500 / 3000 1/min

Rated torque MN 4,1 / 2,5 Nm

Run-up torque MH 7,4 / 5,7 Nm

No-load switching frequency zo 2500 / 1800 s/h

Motor moment of inertia JM 0,00165 kgm2

Brake moment of inertia JBre 0,00007 kgm2

max. braking work until readjustment Wzul. 7 * 107 J

Brake reaktion time t2 0,015 s

(DC-connection)

Braking torque MB 8 Nm

Load torque M

M = P ∗ 9550

nN

ML = 0,74 kW ∗ 9550

2830 1⁄min = 2,5 Nm

Switching torque M U

MU = 2 * MH4

MU = 2 * 7,4 Nm = 14,8 Nm

reduced moment of inertia J red

Jred = 91,2 ∗ m ∗

vnN

2

Jred = 91,2 ∗ (50 kg ∗ 200 kg) ∗

0,24 m⁄s2830 1⁄min

2

= 0,00016 kgm2

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z0 = 2320 / 1620 s/h = max. perm. switching frequency with no load

For this application A 4-2 polemotor (Dahlander-connection) is used. Therefor the half of the z o is criteria.

Permissible switching frequency zzul

up motion: zzul = 1 − ML ⁄ MH

1 + (Jred + JBre) ⁄ JM ∗

zO

2

zzul = 1 − 2,5 Nm ⁄ 5,7 Nm

1 + (0,00016 kgm2 + 0,00007 kgm2) ⁄ 0,00165 kgm2 ∗ 1620 s⁄h

2 = 399 s⁄h (2 poles)

down motion: zzul = 1 − ML ⁄ MU

1 + (Jred + JBre) ⁄ JM ∗

zO

2

zzul = 1 − 2,5 Nm ⁄ 14,8 Nm

1 + (0,00016 kgm2 + 0,00007 kgm2) ⁄ 0,00165 kgm2 ∗ 2320 s⁄h

2 = 846 s⁄h (4 poles)

The mechanical braking depends on the positioning speed. The max. braking distance depends on down motion.

Deceleration a v

av = 9,55 ∗ v ∗ nN4 ⁄ nN2 ∗ (MB − ML ∗ η2)

(Jred ∗ η + JM + JBre) ∗ nN4

av = 9,55 ∗ 0,24 m⁄s ∗ 1400 1⁄min ⁄ 2830 1⁄min ∗ (8 Nm − 2,5 Nm ∗ 0,82)

(0,00016 kgm2 ∗ 0,8 + 0,00165 kgm2 + 0,00007 kgm2) ∗ 1400 1⁄min = 2,80 m⁄s2

In case of calculation the deceleration time we have to use the increased speed for the down motion.

The cause is the delay for switching and the over-synchronous speed.

Load speed n L

nL = nsyn ± ML ⁄ MN ∗ (nsyn − nN) +: down motion, -: up motion

down motion: nL = nsyn + ML ∗ η2 ⁄ MN ∗ (nsyn − nN)

nL = 1500 1⁄min + 2,5 Nm ∗ 0,82

4,1 Nm ∗ (1500 1⁄min − 1400 1⁄min) = 1539 1⁄min

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Increased speed during braking time ∆∆n

∆n = ± 9,55 ∗ ML ∗ t2

Jred + JM + JBre+: down motion, -: up motion

down motion: ∆n = 9,55 ∗ ML ∗ η2 ∗ t2Jred ∗ η + JM + JBre

∆n = 9,55 ∗ 2,5 Nm ∗ 0,82 ∗ 0,015 s

0,00016 kgm2 ∗ 0,8 + 0,00165 kgm2 + 0,00007 kgm2 = 124 1⁄min

Deceleration time t v (Braking time)

tv = v ∗ (nL + ∆n) ⁄ nN2

a

tv = 0,24 m⁄s ∗ (1539 1⁄min + 124 1⁄min) ⁄ 2830 1⁄min

2,80 m⁄s2 = 0,05 s

Deceleration distance sv (Braking distance)

sv = v ∗

nL + ∆nnN2

2

2 ∗ a

sv = 0,24 m⁄s ∗

1539 1⁄min + 124 1⁄min

2830 1⁄min

2

2 ∗ 2,80 m⁄s2 = 0,004 m

Positioning accuracy

The positioning accuracy is about ± 25 % from the deceleration distance s v.

Positioning accuracy = ± 25 % * sv = ± 0,25 * 0,004 m = ± 0,001 m

Braking work WB

WB = (Jred ∗ η + JM + JBre) ∗ n2

N4

182,5 ∗

MB

MB ± ML

WB = (0,00016 kgm2 ∗ 0,8 + 0,00165 kgm2 + 0,00007 kgm2) ∗ (1400 1⁄min)2

182,5 ∗ 8 Nm

8 Nm ± 0 = 20 J

Because of the same number of up- and down-motion the load torque = 0 Nm.

Brake service life until readjustment LN

LN = Wzul

WB ∗ z

LN = 7 ∗ 107J

20 J ∗ 360 1⁄h = 9720 h

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NORD - Information

Gear arrangements

Gear output speed n2

n2 = v ∗ 60π ∗ Dm

n2 = 0,24 m⁄s ∗ 60π ∗ 0,208 m

= 22 1⁄min

^

Mass acceleration factor maf

maf = Jred

JM + JBre

maf = 0,00016 kgm2

0,00165 kgm2 + 0,00007 kgm2 = 0,09

Switching per hour: 1080 ( each 360 accelerations, change-over, decelerations)

⇒ kind of load A, f B = 1,2

Output torque Ma

Ma = PN ∗ 9550

n2 ∗ fB

Ma = 0,75 kW ∗ 9550

22 1⁄min ∗ 1,2 = 391 Nm

Reduction i

i = nN

n2

i = 2830 1⁄min

22 1⁄min = 129

Complete type: SK 2382 A - 80 L/4-2 Bre8PN = 0,60 / 0,75 kWi = 131,86n2 = 11 / 21 1/minMounting position H 1Hollow shaft ø 35 mmBrake 8 NmInsolating material class F

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Example III.1: Turntable drive for processing table

Determine the size of a cd-geared motor for a tuntable with 3 work stations (α = 120°)

Table weight without load m O 500 kg Positioning accuracy = ± 1 mm

Table diameter D 2 m Sprocket reduction i v 3,76

Positions of load α 120° Duty factor ED 60 %

Spaced at radius R 1 m Pulse number 360 Takte/h

Ball bearing ring diameter d 2 m Time of run 16 h/Tag

Cycle time for 120 ° turn t ges 6 s Efficiency η 0,8

Load m L (3 x 750 kg) m L 2250 kg Mounting position V 6

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Distance s ( at a rotation of 120 ° )

s = D ∗ π3

= 2 m ∗ π

3 = 2,094 m

Acceleration time t B or Deceleration time t V

tB = tV = 1 s (acceptance data)

Table speed n T

nT = sges ∗ 60

π ∗ D ∗ ( t − ( tB + tV ) ⁄ 2 ) =

2,094 m ∗ 60π ∗ 2 m ∗ ( 6 s − ( 1 s + 1 s ) ⁄ 2 )

= 4 1⁄ min

Table circumferential velocity v (Ball bearing ring)

v = π ∗ d ∗ nT

60 =

π ∗2 m ∗ 4 1⁄min

60 = 0,42 m⁄s

Moment of inertia J

J = 18

∗ mO ∗ D2 + 14

∗ mL ∗ d2 = 18

∗ 500 kg ∗ (2 m)2 + 14

∗ 2250 kg ∗ (2 m)2 = 2500 kgm 2

Friction power P R (static)

PR = (mO + mL) ∗ g ∗ µ L ∗ v

1000 ∗ η = (500 kg + 2250 kg ) ∗ 9,81 m⁄s2 ∗ 0,005 ∗ 0,42 m⁄s

1000 ∗ 0,8 = 0,07 kW

with µL = 0,005 for friction bearing

Acceleration power P B (dynamic)

PB = J ∗ nT

2

91,2 ∗ 1000 ∗ tB ∗ η = 2500 kgm 2 ∗ (4 1⁄min)2

91,2 ∗ 1000 ∗ 1 s ∗ 0,8 = 0,55 kW

Power P

P = PR + PB (friction + acceleration)

P = 0,07 kW + 0,55 kW = 0,62 kW

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NORD - Information

Motor data

Type 90 S/8-2 Bre 10

Rated power PN 0,25 / 1,1 kW

Rated speed nN 700 / 2810 1/min

Rated torque MN 3,4 / 3,7 Nm

Hochlaufmoment MH 4,0 / 5,7 Nm

No-load switching frequency zo 9000 / 1500 s/h

Motor moment of inertia JM 0,00235 kgm2

Brake moment of inerta JBre 0,00007 kgm2

Braking torque MB (adjusted at 8 Nm) 8 Nm

Load torque ML

M = PR ∗ 9550

nN =

0,07 kW ∗ 95502810 1⁄ min

= 0,2 Nm

Reduced moment of inertia Jred

Jred = J ∗

nT

nN

2 = 2500 kgm2 ∗

4 1⁄min

2810 1⁄min

2 = 0,00507 kgm2

Permissible switching frequence zzul

zzul = 1 − ML ⁄ MH

1 + (Jred + JBre) ⁄ JM ∗ zO = 1 − 0,2 Nm ⁄ 5,7 Nm

1 + (0,00507 kgm2 + 0,00007 kgm2) ⁄ 0,00235 kgm2 ∗ 1500 s⁄h = 453 s⁄h

Acceleration aB

aB = 9,55 ∗ v ∗ (MH − ML)

Jred ⁄ η + JM + JBre) ∗ nN = 9,55 ∗ 0,42 m⁄s ∗ (5,7 Nm − 0,2 Nm)

(0,00507 kgm2 ⁄ 0,8 + 0,00235 kgm2 + 0,00007 kgm2) ∗ 2810 1⁄min = 0,90 m⁄s2

Acceleration time tB (start up time)

tB = vaB

= 0,42 m⁄s0,90 m⁄s2 = 0,47 s

Acceleration distance sB (start up distance)

sB = v2

2 ∗ aB = 0,42 m⁄s2

2 ∗ 0,90 m⁄s2 = 0,098 m

Change-over torque MU

MU = 2 ∗ MH8 = 2 ∗ 4,0 Nm = 8,0 Nm

At the change over the speed increased and the motor is decelet ad generator-style.

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NORD - Information

Change-over delay a U

aU = 9,55 ∗ v ∗ (1 − nN8 ⁄ nN2) ∗ (MU + ML ∗ η2)

(Jred ∗ η + JM + JBre) ∗ (nN2 − nN8) =

aU = 9,55 ∗ 0,42 m⁄s ∗ (1 − 700 1⁄min ⁄ 2810 1⁄min) ∗ (8,0 Nm + 0,2 Nm ∗ 0,82)

(0,00507 kgm2 ∗ 0,8 + 0,00235 kgm2 + 0,00007 kgm2) ∗ (2810 1⁄min − 700 1⁄min) = 1,79 m⁄s2

Change-over time t U

tU = v ∗ (1 − nN8 ⁄ nN2)

aU =

0,42 m⁄s ∗ (1 − 700 1⁄min ⁄ 2810 1⁄min)1,79 m⁄s2 = 0,18 s

Change-over distance s U

sU = (v ∗ (1 − nN8 ⁄ nN2))2

2 ∗ aU =

0,42 m⁄s ∗ (1 − 700 1⁄min ⁄ 2810 1⁄min))2

2 ∗ 1,79 m⁄s2 = 0,028 m

Deceleration a V

aV = 9,55 ∗ v ∗ nN8 ⁄ nN2 ∗ (MB + ML ∗ η2)

(Jred ∗ η + JM + JBre) ∗ nN8 =

9,55 ∗ 0,42 m⁄s ∗ 700 1⁄min ⁄ 2810 1⁄min ∗ (8,0 Nm + 0,2 Nm ∗ 0,82)(0,00507 kgm2 ∗ 0,8 + 0,00235 kgm2 + 0,00007 kgm2) ∗ 700 1⁄min

= 1,80 m⁄s2

Deceleration time tV

tv = v ∗ nN8 ⁄ nN2

aV =

0,42 m⁄s ∗ 700 1⁄min ⁄ 2810 1⁄min

1,80 m⁄s2 = 0,06 s

Deceleration distance s V

sV = (v ∗ nN8 ⁄ nN2 )2

2 ∗ aV =

(0,42 m⁄s ∗ 700 1⁄min ⁄ 2810 1⁄min)2

2 ∗ 1,80 m⁄s2 = 0,003 m

Distance s with velocity v

s = sges - sB - sU - sV = 2,094 m - 0,098 m - 0,028 m - 0,003 m = 1,965 m

Time t with velocity v

t = sv =

1,965 m0,42 m⁄s

= 4,68 s

Pulse duration t ges (total cycle time)

tges = tB + t + tU + tV = 0,47 s + 4,68 s + 0,18 s + 0,06 s = 5,39 s

The required cylce time of 6s is not reached. There is a possibility of driving for a longer period in creep speed.

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NORD - Information

Positioning accuracy

The positioning accuracy is about ± 25% of the deceleration way s V.

Positioning accuracy = ± 0,25 * sv = ± 0,25 * 0,003 m = ± 0,00075 m = ± 0,75 mm

Gear arrangements

Gear output speed n 2

n2 = nT * iV = 4 1/min * 3,76 = 15 1/min

Mass acceleration factor maf

maf = Jred

JM+JBre =

0,00507 kgm2

0,00235 kgm2 + 0,00007 kgm2 = 2,1

Switching per hour: 1080 (each 360 accelerations, change-over and decelerations)

→ kind of load B, f B = 1,5

Output torque M a

Ma = PN ∗ 9550

n2 ∗ fB =

1,1 kW ∗ 955015 1⁄min

∗ 1,5 = 1050 Nm

Reduction i

i = nN

n2 =

2810 1⁄min

15 1⁄min = 187

Complete type: SK 43 - 90 S/8-2 Bre 8PN = 0,25 / 1,1 kWi = 169,86n2 = 4/16 1/minMounting position V6Shaft ø 45 x 90 mmBrake 8 Nm

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